Answer

Verified

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**Hint:**We will consider each of the given data and use it to verify each statement. We need to observe that there could be multiple correct answers. So even if we verified one statement, we need to continue the checking for the remaining statements as well.

**Complete step by step solution:**

It is given that $\hat u = {u_1}\hat i + {u_2}\hat j + {u_3}\hat k$ is a unit vector in ${R^3}$ .

The vector $\hat w = \dfrac{1}{{\sqrt 6 }}\left( {\hat i + \hat j + 2\hat k} \right)$ .

We have already given that there exists a vector $\vec v$ in ${R^3}$ such that $\left| {\hat u \times \vec v} \right| = 1$ and $\hat w\left( {\hat u \times \vec v} \right) = 1$ .

That means we know that there exists at least one vector $\vec v$ , so we only need to check whether it is unique or not.

That will verify the first two statements.

For two vectors $\vec a$ and $\vec b$ the dot product is given by $\vec a \cdot \vec b = ab\cos \theta $ .

It is given that $\hat w\left( {\hat u \times \vec v} \right) = 1$ .

Using the definition of dot product, we write:

$\left| {\hat w} \right|\left| {\left( {\hat u \times \vec v} \right)} \right|\cos \theta = 1$

This implies that $\cos \theta = 1$ .

Therefore, we conclude that the vector $\hat w$ is perpendicular to the cross product vector $\left( {\hat u \times \hat v} \right)$ which will imply that the vector $\hat w$ is perpendicular to both the vectors $\hat u$ as well as $\hat v$ with $\left| {\hat u \times \hat v} \right| = 1$ .

From this we can conclude that there can be infinitely many vectors $\hat v$ satisfying the given conditions.

**Therefore, the option B is definitely correct.**

Now observe that only one of the statements from C and D can be true.

We will begin the verification from statement C.

Let us assume for a moment that the vector $\hat u$ lies in $xy$- plane.

Therefore, the component along $\hat k$ is zero.

The vector $\hat u$ is of the form:

$\hat u = {\hat u_1}i + {\hat u_2}j$

Therefore, the product $\hat w \cdot \hat u = 0$ .

Taking the product by definition we get,

${u_1} + {u_2} = 0$

This implies that $\left| {{u_1}} \right| = \left| {{u_2}} \right|$ .

That means the statement C is also true and statement D is false.

**Therefore, the options B and C both are correct.**

**Note:**The problem is based on the concepts more than the numbers. We don’t even have to make a single calculation just we need to have our concepts clear. Also, we need to keep in mind that the question is such that there could be multiple correct answers so verify each answer before jumping to any conclusion.

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